Thus \(\Sigma \)  is not an unbalanced complete graph. Since the least eigenvalue of the principle submatrix indexed by \(V(H_1)\)  is always less than \(-2\)  and \(n\ge 4\) , then \(V(\Sigma )\)  can be partitioned into four parts \(V_1, V_2, V_3\)  and \(V_4\) , where \(V_1=\lbrace u|uv_2,uv_3\in E(\Sigma )\rbrace \) , \(V_2=\lbrace u|uv_1,uv_3\in E(\Sigma )\rbrace \) , \(V_3=\lbrace u|uv_1,uv_2\in E(\Sigma )\rbrace \)  and \(V_4=\lbrace u|uv_1, uv_2,uv_3\in E(\Sigma )\rbrace \) . Clearly, \(v_1\in V_1\) , \(v_2\in V_2\)  and \(v_3\in V_3\) . Meanwhile, \(V_i\)  is an independent set and every vertex of \(V_j\)  is adjacent to every vertex of \(V_k\)  for \(1\le i\le 3\)  and \(1\le j\ne k\le 4\) . We partition \(V_4\)  to independent sets \(V_{41}\cup \cdots \cup V_{4t}\) , then \(uv\in E\)  for any \(u\in V_{4i}\)  and \(v\in V_{4j}\)  for \(1\le i\ne j\le t\)  and \(t\ge 2\)  through the same methods in [1]}. Therefore, \(\Sigma \)  is a signed completed multipartite graph \((K_{n_1,n_2,\ldots ,n_k},\sigma )\) . If \(\Sigma \)  is balanced, then the proof is completed by Lemmas REF  and REF . Next, let's assume \(\Sigma \)  is unbalanced. Then \(\Sigma \)  must be an unbalanced completed 3-partite graph since \(\Sigma \)  cannot contain an unbalanced \(K_4\)  as an induced subgraph. Let \(V=V_1\cup V_2\cup V_3\) , where \(\mid V_i\mid = n_i\ge 1\)  and \(n=n_1+n_2+n_3\ge 5\) .
